Due to cold weather,a $1\, m$ water pipe of cross-sectional area $1\, cm^2$ is filled with ice at $-10^{\circ}C$. Resistive heating is used to melt the ice. $A$ current of $0.5\, A$ is passed through a $4\, k\Omega$ resistance. Assuming that all the heat produced is used for melting,what is the minimum time required? (In $s$)
(Given: latent heat of fusion for water/ice $= 3.33 \times 10^5\, J/kg$,specific heat of ice $= 2 \times 10^3\, J/(kg\cdot K)$ and density of ice $= 10^3\, kg/m^3$)

Shown in the figure is a semicircular metallic strip that has thickness $t$ and resistivity $\rho$. Its inner radius is $R_1$ and outer radius is $R_2$. If a voltage $V_0$ is applied between its two ends,a current $I$ flows in it. In addition,it is observed that a transverse voltage $\Delta V$ develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is schematic and not drawn to scale)-
$(A)$ $I = \frac{V_0 t}{\pi \rho} \ln \left(\frac{R_2}{R_1}\right)$
$(B)$ the outer surface is at a higher voltage than the inner surface
$(C)$ the outer surface is at a lower voltage than the inner surface
$(D)$ $\Delta V \propto I^2$

The direction of current in an iron-copper thermocouple is

If the temperature of the cold junction decreases,then the neutral temperature

The heater of an electric kettle is made of a wire of length $L$ and diameter $d$. It takes $4$ minutes to raise the temperature of $0.5 \ kg$ of water by $40 \ K$. This heater is replaced by a new heater having two wires of the same material,each of length $L$ and diameter $2d$. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by $40 \ K$?
$(A)$ $4$ if wires are in parallel
$(B)$ $2$ if wires are in series
$(C)$ $1$ if wires are in series
$(D)$ $0.5$ if wires are in parallel

$A$ metallic conductor of irregular cross-section is as shown in the figure. $A$ constant potential difference is applied across the ends $(1)$ and $(2)$. Then: